2460. Apply Operations to an Array

You are given a 0-indexed array nums of size n consisting of non-negative integers.

You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:

• If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.

After performing all the operations, shift all the 0’s to the end of the array.

• For example, the array [1,0,2,0,0,1] after shifting all its 0’s to the end, is [1,2,1,0,0,0].

Return the resulting array.

Note that the operations are applied sequentially, not all at once.

class Solution {
public:
    vector<int> applyOperations(vector<int>& A) {
        int n = A.size();
        vector<int> res;
        for(int i = 0; i < n - 1; i++) {
            if(A[i] == A[i + 1]) {
                A[i] *= 2;
                A[i + 1] = 0;
            }
        }
        for(int i = 0; i < n; i++) {
            if(A[i]) res.push_back(A[i]);
        }
        while(res.size() != n) res.push_back(0);
        return res;
    }
};